OP Malhotra Class 9 Maths Solutions Chapter 18 Surface Area and Volume of 3D Solids Exercise 18 (A)

 

Question 1. Find the surface area of the cuboids whose dimensions are :
(a) 3 m, 4 m, 5 m
(b) 4 cm, 1.7 cm, 2.3 cm
Answer:
(a) For a cuboid with length \( l = 3 \) m, breadth \( b = 4 \) m, and height \( h = 5 \) m:
The formula for the total surface area of a cuboid is \( 2(lb + bh + hl) \).
So, Surface area \( = 2[(3 \times 4) + (4 \times 5) + (5 \times 3)] \)
\( = 2[12 + 20 + 15] \)
\( = 2[47] \)
\( = 94 \) m²
(b) For a cuboid with length \( l = 4 \) cm, breadth \( b = 1.7 \) cm, and height \( h = 2.3 \) cm:
Surface area \( = 2[(4 \times 1.7) + (1.7 \times 2.3) + (2.3 \times 4)] \)
\( = 2[6.8 + 3.91 + 9.2] \)
\( = 2[19.91] \)
\( = 39.82 \) cm²
In simple words: To find the surface area of a cuboid, you need to sum the areas of all its six faces. Think of unfolding the box; the total area of paper you would need is the surface area.

🎯 Exam Tip: Always remember to write the correct units (like m² or cm²) after your final surface area calculation to avoid losing marks.

 

Question 2.
(i) Find the surface area of the cubes whose dimensions are :
(a) 7 cm
(b) 10 m
(ii) To make a model of a cube 96 cm² of board was used. Find the length of one edge of the cube.
(iii) The total surface area of a cube is 726 cm². Find its volume.
Answer:
(i) (a) For a cube with edge \( a = 7 \) cm:
The total surface area of a cube is given by the formula \( 6a^2 \).
Surface area \( = 6 \times (7)^2 \)
\( = 6 \times 49 \)
\( = 294 \) cm²
(b) For a cube with edge \( a = 10 \) m:
Surface area \( = 6 \times (10)^2 \)
\( = 6 \times 100 \)
\( = 600 \) m²
(ii) Given that the total surface area of the cube model is \( 96 \) cm²:
Let the length of one edge of the cube be \( a \).
We know that Surface area \( = 6a^2 \).
So, \( 6a^2 = 96 \)
\( a^2 = \frac{96}{6} \)
\( a^2 = 16 \)
\( a = \sqrt{16} \)
\( a = 4 \) cm
(iii) Given that the total surface area of a cube is \( 726 \) cm²:
Let the length of its edge be \( a \).
Surface area \( = 6a^2 \).
So, \( 6a^2 = 726 \)
\( a^2 = \frac{726}{6} \)
\( a^2 = 121 \)
\( a = \sqrt{121} \)
\( a = 11 \) cm
Now, we need to find the volume of the cube. The formula for the volume of a cube is \( a^3 \).
Volume \( = (11)^3 \)
\( = 11 \times 11 \times 11 \)
\( = 1331 \) cm³
In simple words: For a cube, all sides are equal. The surface area is the area of its six square faces, and the volume is the space it takes up, found by multiplying side by side by side.

🎯 Exam Tip: Remember to distinguish between the formula for surface area (\( 6a^2 \)) and volume (\( a^3 \)) for a cube, and always cube the edge length for volume.

 

Question 3. Complete the following table. The measurements are in cm.

Answer:Let \( l \), \( b \), and \( h \) be the length, width, and height, respectively. The volume \( V \) of a cuboid is given by \( V = l \times b \times h \). We can find the missing values using this formula.
(i) Length \( = 4 \) cm, width \( = 3 \) cm, height \( = 5 \) cm
Volume \( = 4 \times 3 \times 5 = 60 \) cm³
(ii) Length \( = 2 \) cm, width \( = 6 \) cm, height \( = 8 \) cm
Volume \( = 2 \times 6 \times 8 = 96 \) cm³
(iii) Length \( = 7 \) cm, width \( = 3 \) cm, height \( = 4 \) cm
Volume \( = 7 \times 3 \times 4 = 84 \) cm³
(iv) Length \( = 12 \) cm, width \( = 9 \) cm, height \( = 12 \) cm
Volume \( = 12 \times 9 \times 12 = 1296 \) cm³
(v) Length \( = 16 \) cm, width \( = 14 \) cm, height \( = 18 \) cm
Volume \( = 16 \times 14 \times 18 = 4032 \) cm³
(vi) Volume \( = 25984 \) cm³, width \( = 28 \) cm, height \( = 24 \) cm
Length \( = \frac{\text{Volume}}{\text{width} \times \text{height}} = \frac{25984}{28 \times 24} = \frac{25984}{672} = 38.66 \) cm (approximately)
(vii) Volume \( = 2400 \) cm³, length \( = 40 \) cm, width \( = 24 \) cm
Height \( = \frac{\text{Volume}}{\text{length} \times \text{width}} = \frac{2400}{40 \times 24} = \frac{2400}{960} = 2.5 \) cm
(viii) Volume \( = 5400 \) cm³, length \( = 60 \) cm, height \( = 5 \) cm
Width \( = \frac{\text{Volume}}{\text{length} \times \text{height}} = \frac{5400}{60 \times 5} = \frac{5400}{300} = 18 \) cm
The completed table is shown below with all values:In simple words: The volume of a cuboid is found by multiplying its length, width, and height. If you know the volume and two of the sides, you can always find the third side by dividing.

🎯 Exam Tip: Always double-check your calculations, especially when dealing with division or larger numbers, and make sure to use consistent units throughout the problem.

 

Question 4. Find the length of the diagonal of the cuboid whose dimensions are :
(a) 2, 3 and 4 cm
(b) 3, 4 and 5 cm
Answer:
(a) For a cuboid with dimensions length \( l = 2 \) cm, breadth \( b = 3 \) cm, and height \( h = 4 \) cm:
The formula for the length of the diagonal of a cuboid is \( \sqrt{l^2 + b^2 + h^2} \).
Diagonal \( = \sqrt{(2)^2 + (3)^2 + (4)^2} \)
\( = \sqrt{4 + 9 + 16} \)
\( = \sqrt{29} \) cm
\( \approx 5.38 \) cm
(b) For a cuboid with dimensions length \( l = 3 \) cm, breadth \( b = 4 \) cm, and height \( h = 5 \) cm:
Diagonal \( = \sqrt{(3)^2 + (4)^2 + (5)^2} \)
\( = \sqrt{9 + 16 + 25} \)
\( = \sqrt{50} \) cm
\( \approx 7.07 \) cm
In simple words: The diagonal of a cuboid is the longest straight line you can draw inside it, from one corner to the opposite corner. It is calculated using a 3D version of the Pythagorean theorem.

🎯 Exam Tip: Remember to square each dimension, add them, and then take the square root. Be careful with calculations involving square roots, and round to the specified decimal places if required.

 

Question 5. Find the length of the diagonal and the volume of the cubes whose each edge is :
(a) 2 m,
(b) 5 m,
(c) 8 cm.
(ii) Find the volume and the total surface area of a cube if its edge is 12 cm.
Answer:
(i) (a) For a cube with edge \( a = 2 \) m:
The length of the diagonal of a cube is \( \sqrt{3}a \).
Diagonal \( = \sqrt{3} \times 2 = 2\sqrt{3} \) m
\( = 2 \times 1.732 \)
\( = 3.464 \) m
The volume of a cube is \( a^3 \).
Volume \( = (2)^3 = 2 \times 2 \times 2 = 8 \) m³
(b) For a cube with edge \( a = 5 \) m:
Diagonal \( = \sqrt{3} \times 5 = 5\sqrt{3} \) m
\( = 5 \times 1.732 \)
\( = 8.660 \approx 8.66 \) m
Volume \( = (5)^3 = 5 \times 5 \times 5 = 125 \) m³
(c) For a cube with edge \( a = 8 \) cm:
Diagonal \( = \sqrt{3} \times 8 = 8\sqrt{3} \) cm
\( = 8 \times 1.732 \)
\( = 13.856 \approx 13.86 \) cm
Volume \( = (8)^3 = 8 \times 8 \times 8 = 512 \) cm³
(ii) For a cube with edge \( a = 12 \) cm:
Volume \( = a^3 = (12)^3 \)
\( = 12 \times 12 \times 12 = 1728 \) cm³
Total surface area \( = 6a^2 = 6 \times (12)^2 \)
\( = 6 \times 144 = 864 \) cm²
In simple words: For a cube, the diagonal that goes through the center from one corner to the opposite corner is always \( \sqrt{3} \) times the length of its side. This is because all three dimensions are equal.

🎯 Exam Tip: Remember to use the value \( \sqrt{3} \approx 1.732 \) for calculations involving the diagonal of a cube unless specified otherwise. Make sure to clearly state both the diagonal length and the volume for each part.

 

Question 6.
(i) Find the edge of the cubes whose volumes are :
(a) 216 m³
(b) 2197 m³
(ii) Answer true or false. The volume of a rectangular solid measuring 1 m by 50 cm by 0.5 m is 250000 cm³.
(iii) The diagonal of a rectangular solid is \( 5\sqrt{2} \) dm. If its length and breadth are 5 dm and 4 dm, then find the height of the solid.
Answer:
(i) (a) Given volume of a cube \( = 216 \) m³:
To find the edge, we take the cube root of the volume.
Edge \( = \sqrt[3]{216} \)
\( = 6 \) m
(b) Given volume of a cube \( = 2197 \) m³:
Edge \( = \sqrt[3]{2197} \)
\( = 13 \) m
(ii) To check if the statement is true or false, we calculate the volume:
Length \( = 1 \) m \( = 100 \) cm
Breadth \( = 50 \) cm
Height \( = 0.5 \) m \( = 50 \) cm
Volume \( = \text{length} \times \text{breadth} \times \text{height} \)
\( = 100 \times 50 \times 50 \)
\( = 250000 \) cm³
So, the statement is True.
(iii) Given diagonal of rectangular solid \( = 5\sqrt{2} \) dm.
Given length \( l = 5 \) dm and breadth \( b = 4 \) dm.
Let the height be \( h \).
The formula for the diagonal of a cuboid is \( \sqrt{l^2 + b^2 + h^2} \).
So, \( 5\sqrt{2} = \sqrt{(5)^2 + (4)^2 + h^2} \)
Square both sides:
\( (5\sqrt{2})^2 = (5)^2 + (4)^2 + h^2 \)
\( 25 \times 2 = 25 + 16 + h^2 \)
\( 50 = 41 + h^2 \)
\( h^2 = 50 - 41 \)
\( h^2 = 9 \)
\( h = \sqrt{9} \)
\( h = 3 \) dm
In simple words: The edge of a cube is found by reversing the volume calculation – taking the cube root. For cuboids, all dimensions must be in the same units before calculating volume or diagonal.

🎯 Exam Tip: Always ensure all measurements are in the same units (e.g., cm or m or dm) before starting calculations to avoid errors, especially in true/false questions.

 

Question 7. The volume of a rectangular solid is 3600 cm³. If it is 20 cm long and 9 cm high, find its width.
Answer:
Given volume of the rectangular solid \( = 3600 \) cm³.
Given length \( l = 20 \) cm and height \( h = 9 \) cm.
The formula for the volume of a rectangular solid is \( V = l \times b \times h \), where \( b \) is the width.
So, \( 3600 = 20 \times b \times 9 \)
\( 3600 = 180 \times b \)
\( b = \frac{3600}{180} \)
\( b = 20 \) cm
In simple words: If you know how much space a rectangular box takes up (its volume) and how long and tall it is, you can find its width by simply dividing the volume by the length and height multiplied together.

🎯 Exam Tip: Always write down the given values and the formula you are using. This makes it easier to substitute values and solve for the unknown, and helps in showing your steps clearly.

 

Question 8. The length and breadth of a rectangular solid are respectively 25 cm and 20 cm. If the volume is 7000 cm³, find its height.
Answer:
Given length of the rectangular solid \( l = 25 \) cm and breadth \( b = 20 \) cm.
Given volume \( V = 7000 \) cm³.
The formula for the volume of a rectangular solid is \( V = l \times b \times h \), where \( h \) is the height.
So, \( 7000 = 25 \times 20 \times h \)
\( 7000 = 500 \times h \)
\( h = \frac{7000}{500} \)
\( h = 14 \) cm
In simple words: Imagine a rectangular box. If you know how much it can hold (volume) and how long and wide it is, you can figure out how tall it must be.

🎯 Exam Tip: Ensure that you correctly identify which dimension is being asked for and arrange the formula to solve for that specific variable, showing all intermediate calculations.

 

Question 9. The perimeter of one face of a cube is 20 cm. Find (i) the total area of the 6 faces, (ii) the volume of the cube.
Answer:
Given that the perimeter of one face of a cube is \( 20 \) cm.
Each face of a cube is a square. The perimeter of a square is \( 4 \times \text{side} \).
Let the edge (side) of the cube be \( a \).
So, \( 4a = 20 \)
\( a = \frac{20}{4} \)
\( a = 5 \) cm
(i) To find the total area of the 6 faces (total surface area):
The formula for the total surface area of a cube is \( 6a^2 \).
Total surface area \( = 6 \times (5)^2 \)
\( = 6 \times 25 \)
\( = 150 \) cm²
(ii) To find the volume of the cube:
The formula for the volume of a cube is \( a^3 \).
Volume \( = (5)^3 \)
\( = 5 \times 5 \times 5 \)
\( = 125 \) cm³
In simple words: A cube has six identical square faces. If you know the distance around one square face, you can find the length of its side, which then helps you calculate the area of all faces and the total space the cube occupies.

🎯 Exam Tip: Remember that a cube has square faces, so the perimeter of one face refers to the perimeter of a square. Use this to find the side length first before calculating surface area and volume.

 

Question 10. The area of a playground is 4800 m². Find the cost of covering it with gravel 1 cm deep, if the gravel costs Rs. 4.80 per cubic metre.
Answer:
Given area of the playground \( = 4800 \) m².
The depth of gravel to be covered \( = 1 \) cm.
First, convert the depth to metres: \( 1 \) cm \( = \frac{1}{100} \) m \( = 0.01 \) m.
The volume of gravel needed is the area of the base multiplied by the depth.
Volume of gravel \( = \text{Area of playground} \times \text{depth} \)
\( = 4800 \times 0.01 \)
\( = 48 \) m³
The cost of gravel is Rs. \( 4.80 \) per cubic metre.
Total cost \( = \text{Volume of gravel} \times \text{Cost per cubic metre} \)
\( = 48 \times \text{Rs. } 4.80 \)
\( = \text{Rs. } 230.40 \)
In simple words: To find the cost of spreading gravel, first figure out the total amount of gravel needed in cubic metres. Then, multiply this amount by the cost of one cubic metre.

🎯 Exam Tip: Always ensure all units are consistent (e.g., all in metres or all in centimetres) before performing calculations. A common mistake is mixing units like m² and cm in the same calculation.

 

Question 11. A rectangular water tank of base 7 m × 6 m contains water up to a depth of 5 m. How many cubic metres of water are there in the tank?
Answer:
Given the base dimensions of the rectangular water tank are \( 7 \) m \( \times 6 \) m.
The depth of water in the tank is \( 5 \) m.
First, find the area of the base:
Area of the base \( = 7 \times 6 = 42 \) m²
The volume of water in the tank is the area of the base multiplied by the depth of the water.
Volume of water \( = \text{Area of base} \times \text{depth} \)
\( = 42 \times 5 \)
\( = 210 \) m³
In simple words: To find out how much water is in a rectangular tank, you just multiply the length of its base, the width of its base, and the depth of the water. This gives you the total space the water fills.

🎯 Exam Tip: For problems involving volumes of liquids in rectangular containers, the volume is simply the product of the base area and the liquid's depth, provided the units are consistent.

 

Question 12. The internal measurement of a box are 20 cm long, 16 cm wide, and 24 cm high. How many 4 cm cubes could be put into the box?
Answer:
Given the internal dimensions of the box:
Length \( = 20 \) cm
Width \( = 16 \) cm
Height \( = 24 \) cm
First, calculate the volume of the box:
Volume of box \( = \text{length} \times \text{width} \times \text{height} \)
\( = 20 \times 16 \times 24 \)
\( = 7680 \) cm³
Now, consider the smaller cubes, each with an edge (side) of \( 4 \) cm.
Calculate the volume of one such cube:
Volume of one cube \( = (\text{edge})^3 \)
\( = (4)^3 = 4 \times 4 \times 4 = 64 \) cm³
To find how many cubes can be packed into the box, divide the volume of the box by the volume of one cube.
Number of cubes \( = \frac{\text{Volume of box}}{\text{Volume of one cube}} \)
\( = \frac{7680}{64} \)
\( = 120 \)
In simple words: To fit smaller cubes into a big box, you find out how much space the big box has and how much space each small cube takes. Then, you divide the big space by the small space to know how many cubes will fit.

🎯 Exam Tip: This method works perfectly when the dimensions of the larger box are exact multiples of the smaller cube's edge. In other cases, you would have to consider how many cubes fit along each dimension separately.

 

Question 13. The length, breadth and height of a rectangular solid are in the ratio of 5 : 4 : 2. If the total surface area is 1216 cm², find the length, breadth and height of the solid.
Answer:
Given that the length, breadth, and height of a rectangular solid are in the ratio \( 5:4:2 \).
Let the length \( l = 5x \), breadth \( b = 4x \), and height \( h = 2x \) for some value \( x \).
The total surface area is given as \( 1216 \) cm².
The formula for the total surface area of a rectangular solid (cuboid) is \( 2(lb + bh + hl) \).
So, \( 2[(5x)(4x) + (4x)(2x) + (2x)(5x)] = 1216 \)
\( 2[20x^2 + 8x^2 + 10x^2] = 1216 \)
\( 2[38x^2] = 1216 \)
\( 76x^2 = 1216 \)
\( x^2 = \frac{1216}{76} \)
\( x^2 = 16 \)
\( x = \sqrt{16} \)
\( x = 4 \)
Now, we can find the actual dimensions:
Length \( = 5x = 5 \times 4 = 20 \) cm
Breadth \( = 4x = 4 \times 4 = 16 \) cm
Height \( = 2x = 2 \times 4 = 8 \) cm
In simple words: When dimensions are given as a ratio, we use a common multiplier 'x' to represent them. By using the given total surface area, we can solve for 'x' and then find the actual measurements of the box.

🎯 Exam Tip: When dealing with ratios, introduce a variable like 'x' to represent the common factor. This allows you to set up an equation with the given surface area and solve for the variable, leading to the actual dimensions.

 

Question 14. A lock in a canal is 40 m long, 7 m wide. When the sluices are opened, the depth of water in the lock decreases from 5 m to 3 m 80 cm. How many cubic metre of water runs out?
Answer:
Given the length of the canal lock \( l = 40 \) m.
Given the width of the canal lock \( b = 7 \) m.
The initial depth of water is \( 5 \) m.
The final depth of water is \( 3 \) m \( 80 \) cm. We convert \( 80 \) cm to metres: \( 80 \) cm \( = 0.80 \) m.
So, the final depth is \( 3.80 \) m.
The decrease in the depth of water \( h = \text{Initial depth} - \text{Final depth} \)
\( h = 5 \text{ m} - 3.80 \text{ m} \)
\( h = 1.20 \) m
The volume of water that runs out is the volume of a cuboid with the dimensions of the lock and the calculated decrease in height.
Volume of water run out \( = l \times b \times h \)
\( = 40 \times 7 \times 1.20 \)
\( = 280 \times 1.20 \)
\( = 336 \) m³
In simple words: When water flows out of a canal lock, its depth changes. To find the volume of water that left, we calculate the volume of a box whose dimensions are the canal's length and width, and the height is how much the water level dropped.

🎯 Exam Tip: Always be careful with unit conversions. Here, converting centimetres to metres is crucial for correct calculations. Remember to calculate the change in depth accurately before finding the volume.

 

Question 15. How many rectangular blocks, each \( 2\frac{1}{2} \) cm by \( 1\frac{1}{2} \) cm can be packed in a box 90 cm by 78 cm by 42 cm, internal measurements? How many 4 cm cubes can be packed in this box, with faces parallel to the sides?
Answer:
Given internal dimensions of the large box: Length \( L = 90 \) cm, Breadth \( B = 78 \) cm, Height \( H = 42 \) cm.
(i) For rectangular blocks: (Note: The question states "2 1/2 cm by 1 1/2 cm". The solution uses length \( 2\frac{1}{2} \) cm, width \( 2 \) cm, and height \( 1\frac{1}{2} \) cm. We follow the solution's values for consistency in calculation.)
Dimensions of one rectangular block: length \( l = 2\frac{1}{2} \) cm \( = 2.5 \) cm, width \( b = 2 \) cm, height \( h = 1\frac{1}{2} \) cm \( = 1.5 \) cm.
Volume of one block \( = l \times b \times h \)
\( = 2.5 \times 2 \times 1.5 \)
\( = 7.5 \) cm³ (or \( \frac{15}{2} \) cm³)
Volume of the box \( = L \times B \times H \)
\( = 90 \times 78 \times 42 \)
\( = 294840 \) cm³
Number of blocks packed \( = \frac{\text{Volume of box}}{\text{Volume of one block}} \)
\( = \frac{294840}{7.5} \)
\( = 39312 \)
(ii) For 4 cm cubes: (This requires finding how many cubes fit along each dimension, since the cubes must be packed with faces parallel to the sides.)
Edge of one cube \( = 4 \) cm.
Number of cubes along the length of the box \( = \lfloor \frac{L}{\text{edge}} \rfloor = \lfloor \frac{90}{4} \rfloor = \lfloor 22.5 \rfloor = 22 \)
Number of cubes along the breadth of the box \( = \lfloor \frac{B}{\text{edge}} \rfloor = \lfloor \frac{78}{4} \rfloor = \lfloor 19.5 \rfloor = 19 \)
Number of cubes along the height of the box \( = \lfloor \frac{H}{\text{edge}} \rfloor = \lfloor \frac{42}{4} \rfloor = \lfloor 10.5 \rfloor = 10 \)
Total number of 4 cm cubes \( = 22 \times 19 \times 10 \)
\( = 4180 \)
In simple words: To pack items into a box, for the first part, we divide the total space of the big box by the space of one small item. For the second part, since the small cubes must align perfectly, we count how many fit along each side and then multiply these counts.

🎯 Exam Tip: When packing cuboids, always calculate the volume of both the container and the item. For packing identical cubes with parallel faces, calculate how many fit along each dimension (using floor division if needed) and then multiply these counts.

 

Question 16. A tank 72 cm long, 60 cm wide, 36 cm high, contains water to a depth of 18 cm. A metal block 48 cm by 36 cm by 15 cm is put into the tank and totally submerged. Find in cm the amount the water level rises.
Answer:
Given dimensions of the tank: Length \( L_{\text{tank}} = 72 \) cm, Width \( B_{\text{tank}} = 60 \) cm, Height \( H_{\text{tank}} = 36 \) cm.
Initial depth of water \( = 18 \) cm.
The metal block is put into the tank and is totally submerged. This means the volume of water displaced is equal to the volume of the metal block.
Dimensions of the metal block: Length \( l_{\text{block}} = 48 \) cm, Breadth \( b_{\text{block}} = 36 \) cm, Height \( h_{\text{block}} = 15 \) cm.
First, calculate the volume of the metal block:
Volume of block \( = l_{\text{block}} \times b_{\text{block}} \times h_{\text{block}} \)
\( = 48 \times 36 \times 15 \)
\( = 25920 \) cm³
Let the rise in the water level be \( h_{\text{rise}} \). The base area of the tank is \( L_{\text{tank}} \times B_{\text{tank}} \).
The volume of water displaced \( = \text{Area of tank base} \times h_{\text{rise}} \)
So, \( 25920 = (72 \times 60) \times h_{\text{rise}} \)
\( 25920 = 4320 \times h_{\text{rise}} \)
\( h_{\text{rise}} = \frac{25920}{4320} \)
\( h_{\text{rise}} = 6 \) cm
In simple words: When an object is placed in water and sinks completely, it pushes some water out of its way. The amount of water pushed aside is the same as the object's own volume. This displaced water makes the water level in the tank go up.

🎯 Exam Tip: This type of problem relies on the principle of water displacement. The volume of the submerged object equals the volume of the water level rise in the tank, which is the tank's base area multiplied by the height of the rise.

 

Question 17. Find the volume of wood required for making the closed box with internal dimensions : 20 cm by 12.5 cm by 9.5 cm; wood 1.25 cm thick.
Answer:
Given internal dimensions of the closed box: Length \( l_i = 20 \) cm, Breadth \( b_i = 12.5 \) cm, Height \( h_i = 9.5 \) cm.
Given thickness of the wood \( t = 1.25 \) cm.
First, calculate the internal volume of the box:
Internal volume \( V_i = l_i \times b_i \times h_i \)
\( = 20 \times 12.5 \times 9.5 \)
\( = 2375 \) cm³
Now, calculate the external dimensions of the closed box. Since the box is closed, the wood thickness is added to both sides for length, breadth, and height.
External length \( L_o = l_i + 2t = 20 + 2(1.25) = 20 + 2.5 = 22.5 \) cm
External breadth \( B_o = b_i + 2t = 12.5 + 2(1.25) = 12.5 + 2.5 = 15 \) cm
External height \( H_o = h_i + 2t = 9.5 + 2(1.25) = 9.5 + 2.5 = 12 \) cm
Next, calculate the external volume of the box:
External volume \( V_o = L_o \times B_o \times H_o \)
\( = 22.5 \times 15 \times 12 \)
\( = 4050 \) cm³
The volume of wood required is the difference between the external and internal volumes.
Volume of wood \( = V_o - V_i \)
\( = 4050 - 2375 \)
\( = 1675 \) cm³
In simple words: To find how much wood is in a closed box, we imagine a bigger box that includes the wood and a smaller box which is just the empty space inside. The amount of wood is the difference between the volume of the big box and the small box.

🎯 Exam Tip: For a closed box, remember to add twice the wood thickness to each internal dimension (length, breadth, and height) to get the external dimensions. For an open box, only add twice the thickness to length and breadth, and once to height.

 

Question 18. Find the volume of wood required for making the open box with external dimensions 17.5 cm long. 14 cm wide, 10 cm high, wood 7.5 mm thick.
Answer:
Given external dimensions of the open box: Length \( L_e = 17.5 \) cm, Width \( B_e = 14 \) cm, Height \( H_e = 10 \) cm.
Given thickness of the wood \( t = 7.5 \) mm.
First, convert the thickness to cm: \( 7.5 \) mm \( = \frac{7.5}{10} \) cm \( = 0.75 \) cm.
Calculate the external volume of the box:
External volume \( V_e = L_e \times B_e \times H_e \)
\( = 17.5 \times 14 \times 10 \)
\( = 2450 \) cm³
Now, calculate the internal dimensions of the open box. For an open box, the thickness is removed from both sides for length and breadth, but only from the bottom for height (since the top is open).
Internal length \( l_i = L_e - 2t = 17.5 - 2(0.75) = 17.5 - 1.5 = 16 \) cm
Internal breadth \( b_i = B_e - 2t = 14 - 2(0.75) = 14 - 1.5 = 12.5 \) cm
Internal height \( h_i = H_e - t = 10 - 0.75 = 9.25 \) cm
Next, calculate the internal volume of the box:
Internal volume \( V_i = l_i \times b_i \times h_i \)
\( = 16 \times 12.5 \times 9.25 \)
\( = 1850 \) cm³
The volume of wood required is the difference between the external and internal volumes.
Volume of wood \( = V_e - V_i \)
\( = 2450 - 1850 \)
\( = 600 \) cm³
In simple words: For an open box, when you figure out the space inside, remember that the wood takes up space on the bottom and all four sides, but not the top. So, the height inside is only shorter by one layer of wood.

🎯 Exam Tip: The key difference between a closed and an open box in these calculations is how thickness affects the height of the internal dimensions. Always convert all units to be consistent before beginning. (mm to cm, etc.).

 

Question 19. A field is 30 m long and 18 m broad. A pit 6 m long, 4 m wide and 3 m deep, is dug out from the middle of the field and the earth removed is evenly spread over the remaining area to the field, Find the rise in the level of the remaining part of the field in centimetres correct to one decimal place.
Answer:
Given dimensions of the field: Length \( L_{\text{field}} = 30 \) m, Breadth \( B_{\text{field}} = 18 \) m.
Area of the field \( = L_{\text{field}} \times B_{\text{field}} = 30 \times 18 = 540 \) m².
Dimensions of the pit: Length \( l_{\text{pit}} = 6 \) m, Width \( b_{\text{pit}} = 4 \) m, Depth \( h_{\text{pit}} = 3 \) m.
Area of the pit's base \( = l_{\text{pit}} \times b_{\text{pit}} = 6 \times 4 = 24 \) m².
Volume of earth dug out from the pit \( = l_{\text{pit}} \times b_{\text{pit}} \times h_{\text{pit}} \)
\( = 6 \times 4 \times 3 \)
\( = 72 \) m³.
This earth is spread over the remaining area of the field.
Remaining area of the field \( = \text{Area of field} - \text{Area of pit's base} \)
\( = 540 - 24 = 516 \) m².
Let the rise in the level of the remaining part of the field be \( H \).
The volume of earth spread \( = \text{Remaining area} \times H \).
So, \( 72 = 516 \times H \)
\( H = \frac{72}{516} \) m
To convert this height to centimetres:
\( H = \frac{72}{516} \times 100 \) cm
\( H = \frac{7200}{516} \approx 13.953 \) cm
Rounding to one decimal place, the rise in level \( = 13.9 \) cm.
In simple words: When you dig a hole and spread the dirt around, the ground level goes up in the area where the dirt is spread. To find out how much it rises, you divide the volume of the dug-out dirt by the area it covers.

🎯 Exam Tip: Remember to subtract the area of the pit from the total field area to get the correct area over which the earth is spread. Also, pay attention to the required units (cm) and decimal places for the final answer.

 

Question 20. An agricultural field is in the form of a rectangle of length 20 m and width 14 m. A pit 6 m long, 3 m wide and 2.5 m deep is dug in the corner of the field and the earth taken out spread uniformaly over the remaining area of the field. Find the extent to which the level of the field has been raised.
Answer:
Given dimensions of the agricultural field: Length \( L_{\text{field}} = 20 \) m, Width \( B_{\text{field}} = 14 \) m.
Area of the field \( = L_{\text{field}} \times B_{\text{field}} = 20 \times 14 = 280 \) m².
Dimensions of the pit: Length \( l_{\text{pit}} = 6 \) m, Width \( b_{\text{pit}} = 3 \) m, Depth \( h_{\text{pit}} = 2.5 \) m.
Area of the pit's base \( = l_{\text{pit}} \times b_{\text{pit}} = 6 \times 3 = 18 \) m².
Volume of earth dug out from the pit \( = l_{\text{pit}} \times b_{\text{pit}} \times h_{\text{pit}} \)
\( = 6 \times 3 \times 2.5 \)
\( = 45 \) m³.
This earth is spread uniformly over the remaining area of the field.
Remaining area of the field \( = \text{Area of field} - \text{Area of pit's base} \)
\( = 280 - 18 = 262 \) m².
Let the rise in the level of the field be \( H \).
The volume of earth spread \( = \text{Remaining area} \times H \).
So, \( 45 = 262 \times H \)
\( H = \frac{45}{262} \) m
To convert this height to centimetres:
\( H = \frac{45}{262} \times 100 \) cm
\( H = \frac{4500}{262} \approx 17.175 \) cm
Rounding to two decimal places, the rise in level \( = 17.18 \) cm.
In simple words: When earth from a pit is spread over the rest of a field, it makes the field level higher. The amount the field level rises is found by dividing the amount of earth by the area it covers.

🎯 Exam Tip: Always be clear about which area the dug-out earth is spread over. If it's the 'remaining area', subtract the pit's base area from the total field area. Ensure final answer is in correct units and rounded as specified.

 

Question 21. A certain quality of wood costs Rs. 250 per m³. A solid cubical block of such wood is bought for Rs. 182.25. Calculate the volume of the block and use the method of factors to find the length of one edge of the block.
Answer:
Given cost of wood \( = \text{Rs. } 250 \) per m³.
Total cost of the cubical block \( = \text{Rs. } 182.25 \).
To find the volume of the block, we divide the total cost by the cost per cubic metre.
Volume of the block \( = \frac{\text{Total cost}}{\text{Cost per m}^3} \)
\( = \frac{182.25}{250} \) m³
\( = 0.729 \) m³
The block is cubical, so its volume is \( a^3 \), where \( a \) is the length of one edge.
\( a^3 = 0.729 \)
To find \( a \), we take the cube root of \( 0.729 \).
\( a = \sqrt[3]{0.729} \)
\( a = \sqrt[3]{(0.9)^3} \)
\( a = 0.9 \) m
We can also express the edge length in centimetres:
\( a = 0.9 \times 100 \) cm
\( a = 90 \) cm
In simple words: First, use the total cost and the cost per cubic metre to find out how much wood there is. Then, because it's a cube, find the number that, when multiplied by itself three times, gives you that volume. This number is the side length.

🎯 Exam Tip: This problem combines cost calculation with volume and cube root. Make sure your division for volume is accurate, and practice identifying perfect cubes or cube roots of decimal numbers.

 

Question 22. A rectangular container, whose base is a square of side 6 cm, stands on a horizontal table and holds water upto 1 cm from the top. When a cube is placed in the water and is completely submerged, the water rises to the top and 2 cm³ of water overflows. Calculate the volume of the cube.
Answer:
First, find the volume of water when the container is filled to the top. The base is a square with side 6 cm, so the base area is \( 6 \times 6 = 36 \text{ cm}^2 \). The water fills up to 1 cm from the top, which means the empty space at the top is 1 cm high. The cube is placed in the water and it overflows by 2 cm³. This overflow indicates the volume of water pushed out by the cube. When the cube is fully submerged, the water level rises to the top, and 2 cm³ of water spills out. This means the volume of the cube is equal to the volume of water needed to fill the container to the top (the 1 cm empty space) plus the volume of water that overflowed. The volume of water that fills the 1 cm space at the top is \( 36 \text{ cm}^2 \times 1 \text{ cm} = 36 \text{ cm}^3 \). So, the volume of the cube is \( 36 \text{ cm}^3 + 2 \text{ cm}^3 = 38 \text{ cm}^3 \).
In simple words: The container has a square bottom of 6 cm by 6 cm. Water filled it up to 1 cm below the top. When a cube was put in, the water filled the last 1 cm of space and then 2 cm³ spilled over. The cube's volume is the empty space plus the spilled water.

🎯 Exam Tip: Remember that when an object is submerged, it displaces a volume of liquid equal to its own volume. If some liquid overflows, that overflow is part of the displaced volume.

 

Question 23. Two cubes, each with 8 cm edge are joined end to end. Find the surface area of the resulting cuboid.
Answer:
When two cubes, each with an 8 cm edge, are joined end to end, they form a cuboid. The new length of this cuboid will be the sum of the edges of the two cubes, which is \( 8 \text{ cm} + 8 \text{ cm} = 16 \text{ cm} \). The breadth and height of the cuboid will remain the same as the edge of a single cube, which is 8 cm. So, for the resulting cuboid: length \( l = 16 \text{ cm} \), breadth \( b = 8 \text{ cm} \), and height \( h = 8 \text{ cm} \). The surface area of a cuboid is given by the formula \( 2(lb + bh + hl) \).
Substituting the values:
Surface area \( = 2(16 \times 8 + 8 \times 8 + 8 \times 16) \)
\( = 2(128 + 64 + 128) \)
\( = 2(320) \)
\( = 640 \text{ cm}^2 \).
In simple words: When you stick two cubes together side-by-side, the length doubles, but the width and height stay the same. Then, you use the cuboid surface area formula with these new measurements to find the total surface area.

🎯 Exam Tip: Always visualize how the shapes combine. Only one dimension (length, breadth, or height) will change when cubes are joined end-to-end; the other two stay the same as the original cube's edge.

 

Question 24. The areas of three adjacent faces of a cuboid are x, y and z. If the volume is V, prove that \( V^2 = xyz \).
Answer:
Let's consider a cuboid with length \( l \), breadth \( b \), and height \( h \). The areas of three adjacent faces are given as \( x, y, \) and \( z \). We can write these areas in terms of \( l, b, \) and \( h \).
Let \( x = l \times h \) (Area of one face)
Let \( y = b \times h \) (Area of an adjacent face)
Let \( z = l \times b \) (Area of the third adjacent face)
The volume of the cuboid is \( V = l \times b \times h \).
We need to prove that \( V^2 = xyz \).
Let's find the product \( xyz \):
\( xyz = (l \times h) \times (b \times h) \times (l \times b) \)
\( = l^2 \times b^2 \times h^2 \)
We can also write this as \( (l \times b \times h)^2 \).
Since \( V = l \times b \times h \), we can substitute \( V \) into the expression:
\( xyz = V^2 \).
This proves the relationship.
In simple words: Imagine a box. If you multiply the areas of three faces that meet at one corner, you get the square of the box's volume. This is because all the length, width, and height terms get squared.

🎯 Exam Tip: When proving algebraic relationships in geometry, always start by defining the basic dimensions (like l, b, h) and expressing the given quantities (like areas x, y, z, and volume V) in terms of these basic dimensions. Then manipulate the expressions to reach the desired proof.

 

Question 25. A metal cube of edge 12 cm is melted and formed into three smaller cubes. If the edges of two smaller cubes are 6 cm and 8 cm, find an edge of the third smaller cube.
Answer:
When a metal cube is melted and recast into smaller cubes, the total volume of the metal remains the same. First, let's find the volume of the original large cube. The edge of the large cube is 12 cm. The volume of a cube is given by the formula \( a^3 \).
Volume of the large cube \( = (12 \text{ cm})^3 = 12 \times 12 \times 12 = 1728 \text{ cm}^3 \).
This volume is then used to form three smaller cubes. We know the edges of two smaller cubes are 6 cm and 8 cm.
Volume of the first smaller cube \( = (6 \text{ cm})^3 = 6 \times 6 \times 6 = 216 \text{ cm}^3 \).
Volume of the second smaller cube \( = (8 \text{ cm})^3 = 8 \times 8 \times 8 = 512 \text{ cm}^3 \).
Let the edge of the third smaller cube be \( a_3 \). Its volume will be \( (a_3)^3 \).
According to the conservation of volume:
Volume of large cube \( = \) Volume of first cube \( + \) Volume of second cube \( + \) Volume of third cube
\( 1728 = 216 + 512 + (a_3)^3 \)
\( 1728 = 728 + (a_3)^3 \)
Now, subtract 728 from both sides to find the volume of the third cube:
\( (a_3)^3 = 1728 - 728 \)
\( (a_3)^3 = 1000 \text{ cm}^3 \)
To find the edge \( a_3 \), take the cube root of 1000:
\( a_3 = \sqrt[3]{1000} \text{ cm} \)
\( a_3 = 10 \text{ cm} \).
The edge of the third smaller cube is 10 cm. This principle is often used in metallurgy and recycling.
In simple words: When you melt a big cube and make smaller ones, the total amount of metal (volume) stays the same. So, the big cube's volume equals the sum of the volumes of all the small cubes. By finding the volumes of the first two small cubes and subtracting them from the big cube's volume, we find the volume of the third cube, and then its side length.

🎯 Exam Tip: The key principle here is the conservation of volume. When a solid is melted and recast, its volume remains unchanged. Be careful with calculations of cubes and cube roots.

 

Question 26. 50 students sit in a class room. Each student requires 9 m² on floor and 108 m³ in space. If the length of the room is 25 m, find its breadth and height.
Answer:
First, we need to calculate the total floor area and total volume required for all 50 students. Each student needs 9 m² of floor space, so for 50 students, the total floor area required is \( 50 \times 9 \text{ m}^2 = 450 \text{ m}^2 \). Each student needs 108 m³ of space (volume), so for 50 students, the total volume required for the room is \( 50 \times 108 \text{ m}^3 = 5400 \text{ m}^3 \).
The length of the room is given as 25 m.
For a rectangular room (a cuboid), the floor area is given by length × breadth. We know the total floor area and the length, so we can find the breadth:
Breadth \( = \frac{\text{Total floor area}}{\text{Length}} = \frac{450 \text{ m}^2}{25 \text{ m}} = 18 \text{ m} \).
The volume of the room is given by length × breadth × height. We know the total volume, length, and breadth, so we can find the height:
Height \( = \frac{\text{Total volume}}{\text{Length} \times \text{Breadth}} = \frac{5400 \text{ m}^3}{25 \text{ m} \times 18 \text{ m}} \).
\( = \frac{5400 \text{ m}^3}{450 \text{ m}^2} \)
\( = 12 \text{ m} \).
So, the breadth of the room is 18 m and the height is 12 m. Ensuring proper ventilation is also important for student well-being.
In simple words: Find the total floor space and total air space needed for all students. Use the total floor space and the given length to find the width. Then, use the total air space, length, and width to find the height of the room.

🎯 Exam Tip: Remember to calculate the total requirements for all students first. Then, use the formulas for area (\( L \times B \)) and volume (\( L \times B \times H \)) of a cuboid to find the missing dimensions, paying attention to units.

 

Question 27. A cardboard sheet is of rectangular shape with dimensions 42 cm × 36 cm. From each corner, a square of side 6 cm is cut off. An open box is made of the remaining sheet. Find the volume of the box.
Answer:
We start with a rectangular cardboard sheet that is 42 cm long and 36 cm wide. When squares of side 6 cm are cut from each of the four corners, the dimensions of the base of the open box will change. The cuts reduce the length and width of the sheet by twice the side of the cut square (once from each end). The cut-out side length becomes the height of the box when the remaining flaps are folded up.
Original length of sheet \( = 42 \text{ cm} \).
New length of the box \( l = 42 \text{ cm} - (2 \times 6 \text{ cm}) = 42 - 12 = 30 \text{ cm} \).
Original breadth of sheet \( = 36 \text{ cm} \).
New breadth of the box \( b = 36 \text{ cm} - (2 \times 6 \text{ cm}) = 36 - 12 = 24 \text{ cm} \).
The height of the box \( h \) will be equal to the side length of the squares cut from the corners, which is 6 cm.
Now, we calculate the volume of this open box using the formula \( V = l \times b \times h \).
Volume \( = 30 \text{ cm} \times 24 \text{ cm} \times 6 \text{ cm} \)
\( = 720 \text{ cm}^2 \times 6 \text{ cm} \)
\( = 4320 \text{ cm}^3 \).
This method is common in packaging design.
In simple words: When you cut squares from the corners of a flat sheet and fold it to make an open box, the original length and width get shorter by twice the cut size. The height of the box becomes the size of the square you cut out. Then, you multiply the new length, width, and height to get the box's volume.

🎯 Exam Tip: For problems involving cutting squares from corners of a sheet, remember that the length and breadth of the base are reduced by twice the side length of the cut square, while the height of the resulting box is simply the side length of the cut square.

 

Question 28. Two cubes, each of volume 343 cm³ are joined end to end. Find the surface area of the resulting cuboid.
Answer:
First, we need to find the edge length of a single cube. The volume of one cube is given as 343 cm³. Since the volume of a cube is \( a^3 \) (where \( a \) is the edge length), we have:
\( a^3 = 343 \text{ cm}^3 \)
To find \( a \), take the cube root of 343:
\( a = \sqrt[3]{343} = 7 \text{ cm} \).
So, the edge of each cube is 7 cm.
When two such cubes are joined end to end, they form a cuboid. The length of this new cuboid will be the sum of the edges of the two cubes, while the breadth and height remain the same as the edge of a single cube.
Length of resulting cuboid \( l = 7 \text{ cm} + 7 \text{ cm} = 14 \text{ cm} \).
Breadth of resulting cuboid \( b = 7 \text{ cm} \).
Height of resulting cuboid \( h = 7 \text{ cm} \).
Now, we find the total surface area of this resulting cuboid using the formula \( 2(lb + bh + hl) \).
Surface area \( = 2(14 \times 7 + 7 \times 7 + 7 \times 14) \)
\( = 2(98 + 49 + 98) \)
\( = 2(245) \)
\( = 490 \text{ cm}^2 \).
Understanding how joining shapes affects dimensions is key in 3D geometry.
In simple words: First, figure out the side length of one cube from its volume. When you put two of these cubes together side-by-side, the length of the new shape (cuboid) will be twice the cube's side, but the width and height will stay the same. Then, use these new measurements in the formula for the surface area of a cuboid.

🎯 Exam Tip: Always calculate the new dimensions of the combined solid correctly before applying surface area or volume formulas. If cubes are joined end-to-end, only one dimension changes (doubles), while the other two stay the same.